# CRANES

```CRANES I
Mechanical Engineering Department
Carlos III University
CRANES
TRANSPORTATION
CRANES I
INTRODUCTION
• Crane: device used to lift or lower
materials in the vertical direction and to
move them horizontally while being
hanged.
• UNE 58-104-87. Part 1. Types of transport
elevators:
–
–
–
–
According to design.
According to movement possibilities.
According to device control.
According to orientation possibilities.
1
CRANES I
INTRODUCTION
• The dynamic structural calculation allows to determine the
stress of the elevator during its operation.
• Phases:
1.
2.
3.
Find the external forces and their combination, that act on the
structure.
Displacement, stress and reaction calculation of each of the
components applying the adequate calculation process.
Verification of the obtained values of elasticity, resistance and
stability.
Nowadays: Finite element programs are used
CRANES I
INTRODUCTION
–
Principal loads acting on the structure for the motionless elevator. The worst
•
•
–
•
•
–
Self weight: crane components weight (set aside operation load)
Accelerations or decelerations
Vertical impacts due to the rollers
•
•
•
•
Accelerations or decelerations
Centrifugal force
Lateral effects due to rolling
Impact effects
–
Loads due to changes in climate:
–
•
•
Wind, snow and temperature effects
Dimensioning of rails and aisles
2
CRANES I
STRUCTURAL CALCULATION: CASE I
• CASE I: Without wind:
–
The next loads are considered: static due to self weight SG, forces
due to service load SL multiplied by the dynamic coefficient ψ
and the two most unfavourable horizontal effects SH, set aside
impact effects, multiplied by an increase factor γc:
γ c ( SG +ψSL +SH )
–
Increasing factor γc [UNE 58132-2] : Depends on the elevator
classification group
Elevator
group
A1
A2
A3
A4
A5
A6
A7
A8
γc
1,00
1,02
1,05
1,08
1,11
1,14
1,17
1,20
CRANES I
STRUCTURAL CALCULATION: CASE I
γ c ( SG +ψSL +SH )
• Dynamic coefficient ψ: takes into account
–
–
–
The accelerations and decelerations in the lifting process.
The vertical impacts due to the rolling in the track.
Ψ=1+ξVL
VL is the lift speed in m/s
ξ is an experimental coefficient obtained by carrying out several tests in
different elevators
3
CRANES I
STRUCTURAL CALCULATION: CASE I
Bridge and gantry crane
Jib crane
CRANES I
STRUCTURAL CALCULATION: TOWER CRANE
G: crane weight
Jib pendants
Jib pendants
Jib
Counterweight
jib
Pi=Q+Pc
Gc: counterweight weight
Mast
It is important to know the
position: its values are usually
specified in points A and B
4
CRANES I
STRUCTURAL CALCULATION: TOWER CRANE
Above structure
The jib pendants are subject to
traction, while the cat head is
subjected to compression, flexure
and shear
Traction forces in the jib ties
T1 =
PB
sen ( β )
T1 =
Gc
T2 =
sen ( α )
ψ ⋅ PB
sen ( β )
Gc
T2 =
sen ( α )
σ1 =
σ2 =
T1
A1
T2
A2
Forces in the cat head are:
V=T1 ⋅ sen ( β ) +T2 ⋅ sen ( α )
H=T1 ⋅ cos ( β ) -T2 ⋅ cos ( α )
M=H ⋅ h
Von Misses
V
σc =
Ap
σf =
τ=
M
Wpf
H⋅m
b ⋅ Iz
σ T = σ 2 +3τ 2
σ T = (σ f + σ c ) 2 +3τ 2
m: first moment of the
principal area
CRANES I
STRUCTURAL CALCULATION: TOWER CRANE
Above structure
The jib is subjected to flexure and
shear forces:
Von Misses
M pl,f =PA ⋅ L3
M pl,f =ψ ⋅ PA ⋅ L3
Vpl,c =PA
Vpl,c =ψ ⋅ PA
ψ ⋅ PA ⋅ L3
σf =
Wpf
ψ ⋅ PA ⋅ m
τ=
b ⋅ Iz
σ T = σ f2 +3τ 2
5
CRANES I
STRUCTURAL CALCULATION: TOWER CRANE
Mast
M f =PB ⋅ L1 − G c ⋅ L 2 + G ⋅ e
Vc =PB + G c + G
The mast is subjected to flexure and
compression forces
Jib tie
Jib tie
Jib
M f =ψ ⋅ PB ⋅ L1 − G c ⋅ L 2 + G ⋅ e
Vc =ψ ⋅ PB + G c + G
Counterweight
jib
Mf
Vc
L2
e
L
L1=L+e
σf =
ψ ⋅ PB ⋅ L1 − G c ⋅ L 2 + G ⋅ e
Wmf
σc =
ψ ⋅ PB + G c + G
Am
Tower
σ T =σ f + σ C
CRANES I
STRUCTURAL CALCULATION: CASE I
γ c ( SG +ψSL +SH )
• Loads due to horizontal movements:
–
–
–
–
Accelerations and decelerations due to translations
movements of the crane
Acceleration or decelerations due to movements of the
Centrifugal force
Lateral effects due to rolling (loads due to obliquity)
6
CRANES I
STRUCTURAL CALCULATION: CASE I
γ c ( SG +ψSL +SH )
• Accelerations or decelerations of movements :
–
Accelerations/decelerations due to translation movements
of the crane
H=
–
a
⋅V
g
The acceleration/deceleration value depends on:
•
•
•
The desired speed
Time to accelerate/decelerate
Usage of the elevator
CRANES I
STRUCTURAL CALCULATION: CASE I
Desired
speed [m/s]
(a)
Low and medium speed with
large travelling
Time to
accelerate
[s]
Acceleration
[m/s2]
(b)
Medium and high speeds (usual
applications)
(c)
High speed with great
accelerations
Time to
accelerate
[s]
Acceleration
[m/s2]
Time to
accelerate
[s]
Acceleration
[m/s2]
4,00
8,00
0,50
6,00
0,67
3,15
7,10
0,44
5,40
0,58
2,50
6,30
0,39
4,80
0,52
2,00
9,10
0,22
5,60
0,35
4,20
0,47
1,60
8,30
0,19
5,00
0,32
3,70
0,43
1,00
6,60
0,15
4,00
0,25
3,00
0,33
0,63
5,20
0,12
3,20
0,19
2,50
0,16
0,40
4,10
0,098
0,25
3,20
0,078
0,16
2,50
0,064
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CRANES I
STRUCTURAL CALCULATION: CASE I
γ c ( SG +ψSL +SH )
• Accelerations or decelerations of the load
movement:
–
Inertia force of the load (with weight W):
F=
–
Wa
=ma
g
Rotational movement:
T: Inertia moment
J: Inertia polar moment =
α: Angular acceleration
T=Jα
∑m d
i
2
i
CRANES I
STRUCTURAL CALCULATION: CASE I
Inertia forces due to rotation:
F=m e ⋅ a
Equivalent mass
me =∑
Tangencial acceleration
2
i i
2
md
D
F=α ⋅ ∑
a=α ⋅ D
mi d i2
D
8
CRANES I
STRUCTURAL CALCULATION: CASE I
γ c ( SG +ψSL +SH )
–
–
Tangential forces between the wheels and the rail track.
Guide forces.
• A simple translation mechanical model is needed:
–
–
n pairs of wheels in line.
p coupled pairs.
Coupled (C)
Individual (I)
Fixed/Fixed
(F/F)
Fixed/Mobile
(F/M)
CRANES I
STRUCTURAL CALCULATION: CASE I
γ c ( SG +ψSL +SH )
• Loads due to centrifugal forces:
–
Effects of the cable inclination
Fc =
WR  πn 
 
g  30 
2
n: Rotating speed
g: gravity acceleration
9
CRANES I
STRUCTURAL CALCULATION: CASE II
• CASE II: Normal operation with service limit
wind
–
effect of service limit wind Sw and, if needed, the
load due to variation in temperature:
γ c ( SG +ψSL +SH ) + Sw
–
CRANES I
STRUCTURAL CALCULATION: CASE II
γ c ( SG +ψSL +SH ) + Sw
• Wind effect
F=A ⋅ p ⋅ Cf
•A is the net surface, in m2, of the considered element, that is, the solid surface
projection over a perpendicular plane in the wind direction.
•Cf is a shape factor, in the wind direction, for the considered element.
•p is the wind pressure, in kN/m2, and it is calculated by means of the following
equation:
-3
2
p = 0, 613 ⋅10 ⋅ vs [kPa]
where vs is the calculated wind speed in m/s
10
CRANES I
STRUCTURAL CALCULATION: CASE II
Cf: Shape factor [UNE 58-113-85]
Type
Simple
elements
Simple
lattice work
Machine
room, etc
Aerodynamic drag coefficient
l/b or l/D
Description
5
10
20
30
40
50
Metal section in L, in U and flat plates
1,3
1,35
1,6
1,65
1,7
1,9
Circular metal sections
in which Dvs &lt; 6 m2/s
in which Dvs ≥ 6 m2/s
0,75
0,60
0,80
0,65
0,90
0,70
0,95
0,70
1,0
0,75
1,1
0,8
1,55
1,40
1,0
0,8
1,75
1,55
1,2
0,9
1,95
1,75
1,3
0,9
2,1
1,85
1,35
1,0
2,2
1,9
1,4
1,0
Square metal sections of more than
350 mm side and rectangular of more
than 250 mm x 450 mm
b/d
≥2
1
0,5
0,25
Flat side metal section
1,7
Circular metal sections
in which Dvs &lt; 6 m2/s
in which Dvs ≥ 6 m2/s
1,2
0,8
Square structures filled (air cannot flow
beneath the structure)
1,1
CRANES I
STRUCTURAL CALCULATION: CASE II
Wind
Aerodynamic coefficient=
Section proportion=
Element length
1
1
= or
Section height facing wind b
D
Section height facing wind
b
=
Section width parallel to wind d
11
CRANES I
STRUCTURAL CALCULATION: CASE II
Speeds and pressures of service wind [UNE 58-113-85]
Type of crane
Wind speed
m/s
Wind pressure
kPa/m2
Cranes which can be protected against
wind and designed exclusively for light
wind (for example, low height cranes
with an easily folding jib to the ground)
14
0,125
Every normal crane installed in exteriors
20
0,25
28,5
0,50
Dock cranes that should continue
operation in case of strong wind
p = 0, 613 ⋅10-3 ⋅ vs2 [kPa]
CRANES I
STRUCTURAL CALCULATION: CASE II
γ c ( SG +ψSL +SH ) + Sw
–
It is not considered
• Temperature effect:
–
–
Only when the elements cannot freely expand
Temperature limit -20 &ordm;C + 45 &ordm;C
12
CRANES I
STRUCTURAL CALCULATION: CASE III
•
CASE III: Elevator subjected to exceptional loads
a) Elevator out of service subjected to maximum wind.
b) Elevator in service subject to impact.
c) Elevator subjected to static and dynamic tests.
CRANES I
STABILITY
Wm d m &gt; Wb d b + W d
13
CRANES I
STABILITY
Wm
Rotation axis
Wm d m + Wf ( d o − d f ) &gt; Wb d b + Wr d
CRANES I
COUNTERWEIGHT CALCULATION
It is usually calculated so that it counteracts the half
of the material moment and the jib moment
Moment:
G c ⋅ d=G ⋅ e+(Pc +
Q
)⋅L
2
M f =(Pc + Qi ) ⋅ L+G ⋅ e-G c ⋅ d
Q
Q

 
M f =(Pc + Qi ) ⋅ L+G ⋅ e-  G ⋅ e+(Pc + ) ⋅ L  =  Qi −  ⋅ L
2
2

 
P=Qi+Pc
Most unfavourable
situations:
Q
M f =- ⋅ L
2
Mf =
Q
⋅L
2
With this type of counterweight the mast, with and without
load, has a uniform solicitation in a favourable way
14
CRANES I
CLASSIFICATION
Crane and elevators classification allows to establish the design of the
structure and of the mechanisms
It is used by manufacturers and clients so that a specific elevator operates within
certain required service conditions.
Elevator classification
Used by the client and the
manufacturer to achieve a
fixed elevators service
conditions
Mechanism classification
It gives the manufacturer
information of how to design and
verify the elevator so that it has
the desired service life for the
operation service conditions
CRANES I
CLASSIFICATION OF THE EQUIPMENT
CLASSIFICATION OF THE EQUIPMENT (standard 58-112-91/1)
Number of cycles of a manoeuvre
A manoeuvre cycle begins when the load is prepared to be lifted and finishes when
the elevator is prepared to lift the next load
• The total number of manoeuvre cycles is the sum of all of the cycles carried
out during the elevator life.
• The user expects that the elevator’s manoeuvre number of cycles is achieved
during its life.
• The total number of manoeuvre cycles has a relationship with the usage
factor:
–
The manoeuvre spectrum has been conveniently divided in 10 usage classes.
15
CRANES I
CLASSIFICATION OF THE EQUIPMENT: TOTAL NUMBER OF
CYCLES
CRANES I
CLASSIFICATION OF THE EQUIPMENT
CLASSIFICACI&Oacute;N OF THE EQUIPMENT (standard 58-112-91/1)
Number of cycles of a manoeuvre
The load condition is the number of times a load is lifted, which is suitable to the
elevator capacity
• Depending on the available information of the number and weight of the
loads to be lifted during the elevator life:
–
–
Lack of indications: manufacturer and client have to achieve an agreement.
If the information is available: the load spectrum coefficient of the elevator
can be calculated.
16
CRANES I
CLASSIFICATION OF THE EQUIPMENT : LOAD LEVEL
 C  P 3 
Kp = ∑  i  i  
 CT  Pmax  
Ci is the mean number of cycles of manoeuvre for each different load level.
CT is the total of the individual cycles for every load level.
Pi are the values of the individual loads characteristic of the equipment service operation.
Pmax is the maximum load that the equipment is authorized to lift (safe working load).
CRANES I
CLASSIFICATION OF THE EQUIPMENT : LOAD LEVEL
17
CRANES I
CLASSIFICATION OF THE COMPLETE EQUIPMENT
CRANES I
CLASSIFICATION OF THE COMPLETE EQUIPMENT
18
CRANES I
CLASSIFICATION OF THE MECHANISM
CLASSIFICACI&Oacute;N OF THE MECHANISMS (standard 58-112-91/1)
Usage of work equipment
It is calculated for the planned service duration in hours
• Maximum service duration can be computed by means of the mean daily
service, in hours, of the number of working days per year and the number of
the planned years of service.
• A mechanism is considered to be on service when it is on movement.
CRANES I
CLASSIFICATION OF THE MECHANISM: USAGE
19
CRANES I
CLASSIFICATION OF THE MECHANISM
CLASSIFICACI&Oacute;N OF THE MECHANISMS (standard 58-112-91/1)
Use of work equipment
The load level is a feature that shows how much a mechanism is subjected to
t
k m =∑  i
 Tt
3
 Pi  

 
 Pmax  
ti is the mean service duration of the mechanism when subjected to individual loads.
Tt iis the sum of the individual durations in all load level
Pi is the individual load level of the mechanism
Pmax is the maximum load applied to the mechanism
CRANES I
CLASSIFICATION OF THE MECHANISM: APPLIED LOAD
20
CRANES I
CLASSIFICATION OF THE MECHANISM
CRANES I
CLASSIFICATION OF THE MECHANISM
21
CRANES I
CLASSIFICATION
Crane with hook:
CRANES I
CLASSIFICATION
Crane with hook: Usage class U5
Lift cylce
Rotation
Lowering
Rotation
tmc=150 s
22
CRANES I
CLASSIFICATION
Total length usage of the machine:
T=
N ⋅ t mc
[h]
3600
tmc= cycle mean length [s]
N = Number of cycles
T=
5 ⋅105 ⋅150
≈ 20835 horas
3600
CRANES I
CLASSIFICATION
For each of the mechanisms it is defined:
αi =
t mechanism
t mc
Rotation
Lowering
Rotation
tmechanism= usage time of the mechanism during one
cycle [s]
tmc= mean duration of one cycle [s]
Lift mechanism
Slewing mechanism
Movement mechanism
23
CRANES I
CLASSIFICATION
Lift mechanism
Rotation
Lowering
Rotation
Slew mechanism
Rotation
Lowering
Rotation
63%
Travelling mechanism
25%
Rotation
Lowering
Rotation
10%
CRANES I
CLASSIFICATION
Total duration of the mechanism in
hours:
αi=0.63
Lift mechanism
Τi=13126 h
Τ7
Slew mechanism
αi=0.25
Τi=5209 h
Travelling mechanism
Τi=2084 h
Τ5
αi=0.10
Τ4
24
CRANES I
ENGINES
• Power calculation:
Lift movements
G ⋅V
Pe = 2
[CV]
4500 ⋅ η
Travelling movements
(G +G ) ⋅ W ⋅ V
Pt = 1 2
[CV]
4500000 ⋅ η
Power
G1: self weight (trolley,span, etc.) [daN]
V: speed [m/min]
η: mechanical efficiency
W: friction coefficient
7 for rolling bearing
20 for friction bearing
CRANES I
ENGINES
Torque needed to accelerate:
Starting torque = resistance torque + acceleration torque
The resistance torque only has to be taken into account for travelling engines
MA = Mw + Mb [daNm]
Mw =
716 ⋅ Pt
[daNm]
n1
n1: engine speed in rpm
ΣGD12: inertia torque sum referred to engine axis
ta: acceleration time:
Lift, cierre cuchara = 2 s
Trolley travelling or bridge crane, rotation = 4 s
Gantry travelling = 6 s
d=
V
[m]
π ⋅ n1
Mb =
∑ GD
2
1
⋅ n1
375 ⋅ t a
[daNm]
Masses moved linearly
GD12 =
(G1 +G 2 ) ⋅ d 2
η
[daNm 2 ]
Rotating mass
GD12 = GD 22
n 22
[daNm 2 ]
n12
V: Mass lineal speed
25
CRANES I
ENGINES
CRANES I
ENGINES
Power needed to overcome wind resistance:
Pv =
S⋅ V
⋅ Fv [CV]
4500 ⋅ η
Fv: wind pressure [daN/m2]
S: surface exposed to wind
To select travelling engine:
Engine power ≥ Pt +Pv [CV]
Max. motor toruque ≥ M w +M b [daNm]
To select lift engine:
Engine power ≥ Pe +Pv [CV]
26
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